I am currently attempting to prove a claim in Hardy's Course of Pure Mathematics and am currently stuck. I was hoping that someone would be able to provide some assistance on how to go about this.
Claim: Given any rational number r and any positive integer $n$, there exists a rational number on either side of $r$ and differing from $r$ by less than $\frac1n$.
My Proof so far: Any rational number can be written as the ratio of two integers. Let $r = \frac pq$
According to the claim, there exists another rational number, which we will call $r'$ that differs from $r$ by less than $\frac 1n$
Therefore, this can be written as: $|r'- r| < \frac 1n$
And this is where I had become stuck at.
Best Answer
Here is a sketch of one way to approach this problem.
Let a rational number $r$ and a tolerance $\frac{1}{n}$ be given. We want to produce another rational number that differs from $r$ by less than $\frac{1}{n}$.
Take any rational number $s$ greater than $r$. Let $d = s - r$ (the distance between $s$ and $r$). Define another rational number $r_1 = \frac{1}{2}(r + s)$. One can show that $r_1$ is indeed a rational number and has distance $\frac{d}{2}$ from $r$.
If $r_1$ is close enough to $r$ (i.e. if $\frac{d}{2}< \frac{1}{n}$), we can stop. Otherwise, define $r_2 = \frac{1}{2}(r + r_1)$. As before, $r_2$ is a rational number, but its distance from $r$ is $\frac{d}{4}$.
Continue producing these new rational numbers via $r_{k+1} = \frac{1}{2}(r + r_k)$ until one of them is close enough to $r$ (which is guaranteed to occur, since $\frac{d}{2^k} < \frac{1}{n}$ for some $k$).